Hi Jyoti,
>From what you and others have written I am going to guess that you are
using Windows, that Windows has hidden the extension on the file you
are trying to read, and the filename is actually:
galenv.gal
I may be wrong, but this problem has beset others before you.
Jim
>>> I tried to
Hi John,
Both the "right" and "include.lowest" arguments are usually useful
when there are values equal to those in "breaks". A value equal to a
break can fall on either side of the break depending upon these
arguments:
> nums<-1:100
> table(cut(nums,breaks=seq(0,100,by=10)))
(0,10] (10,20] (2
Jeff,
Perhaps I was sloppy with my notation:
I want groups
>=0 <10
>=10 <20
>=20<30
..
>=90 <100
In any event, my question remains, why did the four different versions of cut
give me the same results? I hope someone can explain to me the function of
include.lowest and right in the call to c
Hm...,
Should these two versions produce the same solution? Unfortunately and
shame to confess, I don't know much about the colors in R:
myColors <- c("#FF7C00","#00BF40","#00")
Colors=rgb2hsv(col2rgb(myColors))
apply(Colors,1,mean)
h s v
0.2122974 1.000 0.9163
Thanks to William and Bert!
Atte
16.4.2016, 18.56, William Dunlap kirjoitti:
> Since these are color strings, you can use functions in the grDevices
> package (other others) to manipulate them. E.g., you can convert them
> to various color spaces and perhaps use the mean in one of those
> spaces
Have you read FAQ 7.31 recently, John? Your whole premise is flawed. You should
be thinking of ranges [0,10), [10,20), and so on because numbers ending in 0.9
are never going to be exact.
--
Sent from my phone. Please excuse my brevity.
On April 16, 2016 7:38:50 PM PDT, John Sorkin
wrote:
>
I am trying to understand cut so I can divide a list of numbers into 10 group:
0-9.0
10-10.9
20-20.9
30-30.9,
40-40.9,
50-50.9
60-60.9
70-70.9
80-80.9
90-90.9
As I try to do this, I have been playing with the cut function. Surprising the
following for applications of cut give me the exact same
> On Apr 16, 2016, at 2:03 AM, Akhilesh Singh
> wrote:
>
> Dear All,
>
> I have got your core message, that it is my responsibility to determine
> whether any particular function in my version of R satisfies the language
> requirements at the time of your use. Jim Albert and Maria Rizzo mus
> On Apr 16, 2016, at 6:30 AM, peter dalgaard wrote:
>
> Check the string matching functions, e.g. grepl().
Concretely:
merge3[, grepl( "Email|Email\\.x", names(merge3) ) ] # since "." is special in
grepisch patterns.
Which admittedly is a bit redundant since any character value that includ
Thank you Boris. It worked.
On Sun, Apr 17, 2016 at 1:14 AM, Boris Steipe wrote:
> The command:
>
>system("chflags -R nouchg /path/to/your/directory")
>
> ... should unlock your files recursively in the directory. Then proceed with
> file.remove()
>
> B.
>
>
>
> On Apr 16, 2016, at 3:09 PM, C
The command:
system("chflags -R nouchg /path/to/your/directory")
... should unlock your files recursively in the directory. Then proceed with
file.remove()
B.
On Apr 16, 2016, at 3:09 PM, Christofer Bogaso
wrote:
> Hi,
>
> I am looking for some R code, which will delete all files in a
Hi,
I am looking for some R code, which will delete all files in a Folder
that contains both Locked and Unlocked files. There are many,
therefore i would like to delete all files programmatically in one go.
I used following code :
## "SS" is the Folder name including entire path which contains L
See here:
http://www.r-bloggers.com/search/twitter
On Apr 16, 2016, at 2:58 PM, Ragia . wrote:
>
> Dear group,
> I want to download text of weets and other data such as user id or so for a
> specific hashtag on twitter, how to do this..any reference suggestions that
> could help me to down
Dear group,
I want to download text of weets and other data such as user id or so for a
specific hashtag on twitter, how to do this..any reference suggestions that
could help me to download almost it all without limits in number of dlwonloaded
tweets.
thanks in advance
RaGiA
On 16/04/2016 12:33 PM, Atte Tenkanen wrote:
Hm...,
Should these two versions produce the same solution?
I wouldn't expect them to.
Duncan Murdoch
Unfortunately and
shame to confess, I don't know much about the colors in R:
myColors <- c("#FF7C00","#00BF40","#00")
Colors=rgb2hsv(col2
grDevices has `convertColor()` and the `colorspace` has other
functions that can convert from RBG to Lab space. You should convert
the RGB colors to Lab and average them that way (or us other functions
to convert to HSL or HSV). It all depends on what you are trying to
accomplish with the "average"
On 16/04/2016 8:47 AM, Atte Tenkanen wrote:
Hi,
How would you calculate the "mean colour" of several colours, for
example c("#FF7C00","#00BF40","#00")?
Bert answered your subject line question. Your text is asking something
else: if those are colours, you don't want to treat each of th
Since these are color strings, you can use functions in the grDevices
package (other others) to manipulate them. E.g., you can convert them
to various color spaces and perhaps use the mean in one of those
spaces as your 'average color'.
> myColors <- c(One="#FF7C00",Two="#00BF40",Three="#00
... and if you need to convert back: ?as.hexmode
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Apr 16, 2016 at 8:20 AM, Bert Gunter wrote:
>
?strtoi
You'll have to remove the "#" first, e.g. via substring()
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Apr 16, 2016 at 5:47 AM, Atte
Hi,
How would you calculate the "mean colour" of several colours, for
example c("#FF7C00","#00BF40","#00")?
Yours,
Atte Tenkanen
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
P
You neglected the commas in your index expression!
See ?match
as in:
> out <- cost[match(data,index)]
> out
[1] 50 100 NA
(and "data" is a bad name to use as there is already a data() function in R).
Please DO go through an R tutorial or two to learn about some of these
basic, useful R capa
On Sat, 16 Apr 2016, ch.elahe via R-help wrote:
-Thank you James, well the problem of my type of data is that there can
be many possible subsets and therefore plots, and I want to
automatically generate them, and facet_wrap does not give me all the
possible cases
Not true.
It may get crampe
Use the split function to automatically create a list of pre-subsetted
data frames, and then generate your output however you wish to. For
example (using Jim Lemon's sample data generator):
library(ggplot2)
mydata <- data.frame( RE = sample( 5:50, 100, TRUE)
, LU = sample(
Hi Dominik
Welcome to R-help
Please have a look at
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and/or http://adv-r.had.co.nz/Reproducibility.html for some suggestions on how
to ask questions here.
In particular we probably need to know what packages ha
-Thank you James, well the problem of my type of data is that there can be many
possible subsets and therefore plots, and I want to automatically generate
them, and facet_wrap does not give me all the possible cases
On Saturday, April 16, 2016 6:01 AM, James C. Whanger
wrote:
Would facet
Check the string matching functions, e.g. grepl().
-pd
> On 16 Apr 2016, at 15:18 , Dan Abner wrote:
>
> Hi all,
>
> I want to select all variables in the data.frame with a name that
> includes are certain string. Something like the following:
>
> merge3[,names(merge3) %in% c("Email","Email.x
Hi all,
I want to select all variables in the data.frame with a name that
includes are certain string. Something like the following:
merge3[,names(merge3) %in% c("Email","Email.x")]
But there are too many variations on the Email variable names to list them all.
Can anyone advise?
Thanks!
Dan
Dear all,
I am trying to simulate a series of networks that have characteristics
similar to real life social networks. Specifically I am interested in
networks that have (a) a reasonable degree of clustering (as measured by
the transitivity function in igraph) and (b) a reasonable degree of degree
Would facet_wrap or facet_grid give you what you want?
On Sat, Apr 16, 2016 at 8:45 AM, ch.elahe via R-help
wrote:
> Hi,
> I have a data set (mydata), which a part of this is like the following:
>
>
> 'data.frame': 36190 obs. of 16 variables:
> $ RE: int 38 41 11 67 30 18
Hi,
I have a data set (mydata), which a part of this is like the following:
'data.frame': 36190 obs. of 16 variables:
$ RE: int 38 41 11 67 30 18 38 41 41 30 ...
$ LU : int 4200 3330 530 4500 3000 1790 4700 3400 3640
4000 ...
$ COUNTRY: Fa
Dear Dominik,
There's not enough information here to answer your question: You don't show the
commands you used, nor the output that was produced, and you don't share your
data. A good guess, however, is that there is only one level of the factor,
perhaps after missing data are removed, and so
Hi tan sj,
It is by no means easy to figure out what you want without the code,
but If I read your message correctly, you can run the loops either
way. When you have nested loops producing output, it is often a good
idea to include the parameters for each run in the output as well as
the result so
Dear community
I'm pretty new to R and I'm trying to do a Point-baserial correlation for a
nominal dichotomous variable with a interval scaled variable. It works fine,
but the output just shows me the correlation and nothing else (p-Value would be
important).
I tried it with the following c
Dear All,
I have got your core message, that it is my responsibility to determine
whether any particular function in my version of R satisfies the language
requirements at the time of your use. Jim Albert and Maria Rizzo must have
used their code, which was permitted in the R-code of their time (2
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